Given the parabola
$ 9 x^2 - 12 x y + 4 y^2 + 2 x + 8 y - 10 = 0 $
Find its shortest distance to the line $ x - 2 y + 8 = 0 $
My attempt:
First, we have to make sure that the line does not intersect the parabola, so substituting the line into the parabola we get
$ 4 (2y-8)^2 - 4 y (2y-8) + y^2 + 2 (2y-8) + 8 y - 10 = 0 $
which simplifies to
$ 25 y^2 - 84 y + 230 = 0 $
Discriminant : $(84)^2 - 4(25)(230) \lt 0 $
At the shortest distance, the gradient of the parabola is along the normal to line, so
$ [ 18 x - 12 y + 2 , - 12 x + 8 y + 8 ] = \alpha [1, -2] $
Eliminating $\alpha$ , we obtain
$ -2 (18 x - 12 y + 2) - (-12 x + 8 y + 8 ) = 0 $
So that,
$ - 24 x + 16 y - 12 = 0 $
Divide by $4$
$ - 6 x + 4 y - 3 = 0 $
or
$ y = \dfrac{1}{4} (6 x + 3) $
Substitute this into the equation of the parabola
$ 9 x^2 - 3 x (6 x + 3) + \dfrac{1}{4} (6x + 3)^2 + 2 x + 2 (6 x + 3) - 10 = 0 $
Multiply through by $4$
$ 36 x^2 - 12 x (6x + 3) + (6x + 3)^2 + 8 x + 8 (6 x + 3) - 40 = 0 $
Simplify,
$ 56 x - 7 = 0 $
Therefore, $ x = \dfrac{1}{8} $
$ y = \dfrac{1}{4} (6 x + 3) = \dfrac{1}{4} (\dfrac{15}{4} ) = \dfrac{15}{16} $
Now, we only need to find the distance of $( \dfrac{1}{8} , \dfrac{15}{16} ) $ from the line $ x - 2 y + 8 = 0 $
$ d = \dfrac{ | \dfrac{1}{8} - \dfrac{15}{8} + 8 | }{ \sqrt{5} } = \dfrac{5 \sqrt{5}}{4} $
My Question: Is the method that I used correct, and are my findings correct ? Any feedback is greatly appreciated.